So I'm working on a nuclear physics problem and am looking at radioactive decay. The common unit used for very long decays is years within the literature. Is this the sidereal or tropical year? I want to use units of seconds but seeing as how these 20 minutes 24.5 seconds that differential will add up over time...

I would guess tropical but that's just a guess.

And on the same note, what about days? 24 hours? or 23 hours 56 minutes and 4.1 seconds?

Bonus points for a source

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    $\begingroup$ This reminds me of a joke : Some tourists visit the great pyramids in Giza. The guide tells them that the pyramids are 4504.5 years old. Tourists are impressed by this precision, and ask the guide how it was calculated : "That's very simple, I've been working for 4 and a half years here, and the pyramids were 4500 years old when I started". $\endgroup$ Commented Apr 24, 2018 at 7:07
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    $\begingroup$ How big is your error? $\endgroup$
    – jpmc26
    Commented Apr 24, 2018 at 7:46
  • $\begingroup$ For most of the cases the error in determining the halflife on big year scales is bigger than the one you introduce by using "the wrong" year length to calculate it back. $\endgroup$
    – PlasmaHH
    Commented Apr 24, 2018 at 9:09
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    $\begingroup$ Your question switched from years to days. The difference between sidereal and solar days is about 1/365.25, but the difference between sidereal and tropical years is about 4e-5, or 1/26000, and the reason for the difference is unrelated. $\endgroup$ Commented Apr 24, 2018 at 12:35
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    $\begingroup$ I recall, back around 1966, a college prof who liked to approximate it as pi * 10**7. $\endgroup$
    – Hot Licks
    Commented Apr 26, 2018 at 1:59

4 Answers 4


A "year" without qualification may refer to a Julian year (of exactly $31\,557\,600~\rm s$), a mean Gregorian year (of exactly $31\,556\,952~\rm s$), an "ordinary" year (of exactly $31\,536\,000~\rm s$), or any number of other things (not all of which are quite so precisely defined).

Radioactive decay tables tend to be compiled from multiple different sources, most of which don't clarify which definition of "year" they used, so it is unclear what definition of year is used throughout. It's quite possible that many tables aren't even consistent with the definition of "year" used to calculate the decay times.

On the other hand, the standard error is usually overwhelmingly larger than the deviation created by using any common definition of year, so it doesn't really make a difference.

A day in physics without qualification pretty universally refers to a period of exactly $86\,400~\rm s$.


Years are merely an approximation, as you pointed out, they really aren't precisely defined. In physics seconds are used as they can be calculated exactly using atomic clocks.

For instance, no application requires an exact decimal representation of years, you can round to approximate numbers and then use a remainder of seconds.

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    $\begingroup$ U-238 is listed as $4.468 \pm 0.005 \times 10^9$ years, while any systematic error from the definition of years should top out at $\pm 0.0004 \times 10^9$ years. $\endgroup$
    – JEB
    Commented Apr 24, 2018 at 3:40
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    $\begingroup$ @JEB could you share your math for that? my attempts to replicate it aren't quite matching up. $\endgroup$
    – Tin Wizard
    Commented Apr 24, 2018 at 19:49
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    $\begingroup$ @Walt I was winging it, so let's see. TwitchyKid said he had a 20 minute uncertainty (peak-to-peak) in his year definition, which 1 part in 3 in 24 in 365, or 1 part in 26000. Multiply that by 4.5 billion and I get 0.0002 billion. Which is not what I said. I better check my .python_history file. $\endgroup$
    – JEB
    Commented Apr 24, 2018 at 21:31
  • $\begingroup$ @JEB After $10^9$ years of 20 minutes' error each, you could expect ~40,000 years of error maximum. I think you got it about right. (or possibly three orders of magnitude off; I can't read scientific notation) $\endgroup$
    – anon
    Commented Apr 25, 2018 at 18:53

As a not necessarily representative example, the decay data in the NUCLIDES 2000 database, which is based on the JEF2.2 decay data file, use a year of 365 days and a day of 24 hours.

For example, the halflife of Co-60 is internally stored as 1.6623E+08 seconds but reported as 5.2711E+00 years.

The decay data provided in

  • Endo, A., Yamaguchi, Y., Eckerman, K.F., 2005. Nuclear Decay Data for Dosimetry Calculations: Revised Data of ICRP Publication 38. JAERI 1347. Japan Atomic Energy Research Institute, Tokaimura, Naka-gun, Ibaraki.
  • ICRP, 2008. Nuclear Decay Data for Dosimetric Calculations. ICRP Publication 107. Ann. ICRP 38 (3).

use a year of 365.2422 days and a day of 24 hours (which corresponds to a tropical year of 365 days, 5 hours, 48 minutes, and 46 seconds).


As stated in previous answers, the difference between the different year definitions is too small to matter unless you are dealing with very precise measurements. However, it seems that some people do have data precise enough to care about an exact definition of the year.

In astronomy, when the term year is used as a unit of time (rather than a variable astronomical period), it is understood as a Julian year of exactly 365.25 × 86400 SI seconds. This is the basis of the definition of the light-year.

This definition, however, does not necessarily apply to other disciplines. In 2011, the International Union of Pure and Applied Chemistry (IUPAC) and the International Union of Geological Sciences (IUGS) jointly recommended to define the year (called annus, symbol: a) as 3.1556925445e7 SI seconds, which is the length of the tropical year for the epoch 2000.0. The recommendation seemed to meet some resistance, so do not take for granted that everybody follows it.

My guess is that anyone publishing data precise enough for the definition to really matter will presumably specify the precise definition they are using.


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